Graph transformations: translations and reflections

Graph transformations produce new graphs from familiar ones. Instead of making a fresh table of values, you can move every point of $y=f(x)$ according to one precise rule.

The essential distinction is this:

• a change outside $f(x)$ changes the output, so it acts vertically;
• a change inside the brackets changes the input, so it acts horizontally.

The horizontal direction is the main conceptual hurdle. In $f(x+3)$, the graph moves left, not right. This page explains why.

Prerequisites

You should be able to:

• plot and read coordinates;
• identify roots, intercepts and turning points;
• substitute into an expression;
• recognise basic graphs such as $y=x^2$, $y=x^3$ and $y=1/x$;
• understand that $f(x)$ is the output of a function at input $x$.

Review function notation and graphs of common functions if these ideas are not yet secure.

The four rules

Starting from $y=f(x)$, where $a>0$:

New graph	Transformation	Coordinate rule
$y=f(x)+a$	translate up by $a$	$(x,y)\mapsto(x,y+a)$
$y=f(x)-a$	translate down by $a$	$(x,y)\mapsto(x,y-a)$
$y=f(x-a)$	translate right by $a$	$(x,y)\mapsto(x+a,y)$
$y=f(x+a)$	translate left by $a$	$(x,y)\mapsto(x-a,y)$
$y=-f(x)$	reflect in the $x$-axis	$(x,y)\mapsto(x,-y)$
$y=f(-x)$	reflect in the $y$-axis	$(x,y)\mapsto(-x,y)$

A translation is also described by a vector. For example, right by $4$ and down by $2$ is

$$\begin{pmatrix}4\\-2\end{pmatrix}.$$

Translations and reflections preserve shape. They change a graph’s position or orientation, not its basic form.

Vertical translations: changes outside the function

In

$$y=f(x)+a,$$

$a$ is added to every output. The $x$-coordinate stays fixed while the $y$-coordinate changes:

$$(x,y)\mapsto(x,y+a).$$

Worked example 1: move a quadratic vertically

Sketch $y=x^2-4$ from $y=x^2$.

Here $f(x)=x^2$ and the change is outside the function:

$$y=f(x)-4.$$

Every point moves down by $4$:

$$\begin{aligned} (0,0)&\mapsto(0,-4),\\ (1,1)&\mapsto(1,-3),\\ (-1,1)&\mapsto(-1,-3). \end{aligned}$$

The vertex moves from $(0,0)$ to $(0,-4)$, so the axis of symmetry remains $x=0$.

The roots do not come from moving the old root down, because a root must lie on the $x$-axis. Solve the new equation:

$$x^2-4=0$$ $$(x-2)(x+2)=0,$$

so the roots are $x=-2$ and $x=2$.

Self-check 1

The point $(-3,5)$ lies on $y=f(x)$. What point lies on $y=f(x)+7$?

Answer

The output increases by $7$, while the input is unchanged:

$(-3,5)\mapsto(-3,12).$

Horizontal translations: changes inside the function

In

$$y=f(x-a),$$

the graph moves right by $a$. The sign appears to point the wrong way because the expression changes the input, not the output.

Suppose $(u,v)$ lies on $y=f(x)$. Then

$$v=f(u).$$

To obtain the same output from $f(x-a)$, its input must still equal $u$:

$$x-a=u.$$

Therefore $x=u+a$. The old point $(u,v)$ appears $a$ units to the right:

$$(u,v)\mapsto(u+a,v).$$

Similarly, $f(x+a)$ moves the graph left by $a$.

Worked example 2: move a quadratic horizontally

Sketch $y=(x-3)^2$ from $y=x^2$.

Writing $f(x)=x^2$ gives

$$(x-3)^2=f(x-3).$$

Therefore the graph translates $3$ units right:

$$\begin{aligned} (0,0)&\mapsto(3,0),\\ (1,1)&\mapsto(4,1),\\ (-1,1)&\mapsto(2,1). \end{aligned}$$

The vertex is $(3,0)$ and the axis of symmetry is $x=3$.

There is a useful algebra check. The minimum occurs when the squared expression is zero:

$$x-3=0\quad\Longrightarrow\quad x=3.$$

Worked example 3: track a feature rather than individual points

The graph $y=f(x)$ has a root at $x=-2$ and a turning point at $(4,7)$. State the corresponding features of

$$y=f(x+5).$$

The $+5$ is inside the function, so the entire graph moves left by $5$:

$$(x,y)\mapsto(x-5,y).$$

Hence

$$(-2,0)\mapsto(-7,0)$$

and

$$(4,7)\mapsto(-1,7).$$

The new root is $x=-7$ and the new turning point is $(-1,7)$.

Self-check 2

The point $(6,-1)$ lies on $y=g(x)$. Find the corresponding point on $y=g(x-4)$.

Answer

$g(x-4)$ moves the graph right by $4$:

$$(6,-1)\mapsto(10,-1).$$

Check: at the new input $x=10$, the function receives $10-4=6$.

Combining horizontal and vertical translations

For

$$y=f(x-a)+b,$$

the graph moves right by $a$ and up by $b$:

$$\boxed{(x,y)\mapsto(x+a,y+b)}.$$

The translation vector is

$$\begin{pmatrix}a\\b\end{pmatrix}.$$

The signs in the final equation must be read carefully: the horizontal sign is reversed, while the vertical sign is not.

Worked example 4: translate a reciprocal graph

Describe the transformation from

$$y=\frac1x$$

to

$$y=\frac1{x+2}-3.$$

Let $f(x)=1/x$. Then the new graph is

$$y=f(x+2)-3.$$

It moves left by $2$ and down by $3$, with translation vector

$$\begin{pmatrix}-2\\-3\end{pmatrix}.$$

The original asymptotes $x=0$ and $y=0$ move with the graph, becoming

$$x=-2 \qquad\text{and}\qquad y=-3.$$

The expression is undefined at $x=-2$, which confirms the vertical asymptote.

Worked example 5: form an equation from a translation

The graph $y=x^2$ is translated by

$$\begin{pmatrix}5\\-2\end{pmatrix}.$$

Find its new equation.

The graph moves right by $5$, so replace $x$ with $x-5$. It moves down by $2$, so subtract $2$ outside:

$$\boxed{y=(x-5)^2-2}.$$

The vertex $(0,0)$ moves to $(5,-2)$, which provides a quick check on both signs.

Reflection in the x-axis

For

$$y=-f(x),$$

every output changes sign:

$$(x,y)\mapsto(x,-y).$$

Points on the $x$-axis remain fixed. Points above it move the same distance below it, and vice versa.

Worked example 6: reflect a cubic vertically

Reflect $y=x^3-2$ in the $x$-axis.

The whole output must be negated:

$$\begin{aligned} y&=-(x^3-2)\\ &=-x^3+2. \end{aligned}$$

The brackets matter. Writing $-x^3-2$ would negate only the $x^3$ term and would not reflect the complete graph.

For example, $(0,-2)$ maps to $(0,2)$, as the reflected equation confirms.

Reflection in the y-axis

For

$$y=f(-x),$$

every input changes sign:

$$(x,y)\mapsto(-x,y).$$

Points on the $y$-axis remain fixed.

Worked example 7: distinguish the two reflections

Let

$$f(x)=x^2+3x.$$

Reflection in the $x$-axis gives

$$\begin{aligned} y&=-f(x)\\ &=-(x^2+3x)\\ &=-x^2-3x. \end{aligned}$$

Reflection in the $y$-axis gives

$$\begin{aligned} y&=f(-x)\\ &=(-x)^2+3(-x)\\ &=x^2-3x. \end{aligned}$$

Outside $f$ negates the output. Inside $f$ negates the input. These are different operations.

Symmetry can hide a reflection

For $f(x)=x^2$,

$$f(-x)=(-x)^2=x^2=f(x).$$

The graph is symmetric about the $y$-axis, so reflecting it in that axis makes no visible change. The transformation rule still applies.

Self-check 3

The graph $y=h(x)$ passes through $(-4,3)$.

1. Which point lies on $y=-h(x)$?
2. Which point lies on $y=h(-x)$?

Answers

1. Reflection in the $x$-axis changes the output: $(-4,3)\mapsto(-4,-3)$.
2. Reflection in the $y$-axis changes the input: $(-4,3)\mapsto(4,3)$.

A reliable sketching method

When transforming a graph:

1. identify whether each change is inside or outside $f$;
2. state the transformation in words or as a coordinate rule;
3. mark important points such as roots, intercepts and turning points;
4. transform those points and any asymptotes;
5. join them with the same basic graph shape;
6. check one point by substitution into the new equation.

Do not simply move the axes. The coordinate axes stay fixed while the graph moves.

Common misconceptions

• Reading $f(x+4)$ as right by $4$. It moves left by $4$. Solve $x+4=u$ to see that the old input $u$ occurs at $x=u-4$.
• Changing the wrong coordinate. Outside changes affect $y$; inside changes affect $x$.
• Moving only the vertex. Every point, intercept and asymptote follows the same rule.
• Treating $-f(x)$ and $f(-x)$ as identical. They reflect in different axes.
• Negating only one term. If $f(x)=x^2-5x+1$, then $-f(x)=-x^2+5x-1$.
• Assuming intercepts keep their roles. A transformed old point remains on the graph, but it may no longer be an intercept. Find new intercepts from the transformed equation when needed.

Mixed self-check

Question 1

Describe fully the transformation from $y=f(x)$ to

$$y=f(x-6)+1.$$

Question 2

The point $(2,-5)$ lies on $y=f(x)$. Find its image on

$$y=-f(x+3)+4.$$

Question 3

Write the equation obtained by reflecting $y=2x^2-3x+4$ in the $y$-axis.

Question 4

The graph $y=f(x)$ has vertical asymptote $x=1$ and horizontal asymptote $y=-2$. State the corresponding asymptotes of

$$y=f(x-4)+7.$$

Answers

1. Translate right by $6$ and up by $1$, with vector

$$\begin{pmatrix}6\\1\end{pmatrix}.$$

2. First use the input condition. The $x+3$ moves the point left by $3$:

$$(2,-5)\mapsto(-1,-5).$$

Then $-f$ negates the output and $+4$ raises it:

$$-(-5)+4=9.$$

Therefore the image is $(-1,9)$.

3. Replace every $x$ by $-x$:

$$\begin{aligned} y&=2(-x)^2-3(-x)+4\\ &=2x^2+3x+4. \end{aligned}$$

4. The entire graph moves right by $4$ and up by $7$. Therefore

$$x=1+4=5$$

and

$$y=-2+7=5.$$

What to remember

$$\boxed{\text{Outside }f:\text{ vertical and direct}}$$ $$\boxed{\text{Inside }f:\text{ horizontal and opposite}}$$

In particular,

$$f(x)+a:\text{ up}, \qquad f(x+a):\text{ left},$$ $$-f(x):\text{ reflect in the }x\text{-axis}, \qquad f(-x):\text{ reflect in the }y\text{-axis}.$$

Next, consolidate the parent shapes in graphs of common functions. Then study A-level graph transformations for stretches, compressions, modulus graphs and more demanding combined transformations. Function graphs develops intercepts, asymptotes and symmetry further.